IMPULSE AND MOMENTUM 259
205. Case II. Rough Contact. When the plane is rough frictional forces come into play and change the tangential component of the momentum. Let F be the tangential force due to friction, N the normal force, and & the coefficient of friction; theu we have
X
= -mvn,
'= / Ndt = -JT
Lt = / F dt I fjN dt =
JQ JQ
f*T' f*T'
L/= I Fdt= I vN dt = efj,mvn. JT JT
But Lt+ L^ = mo I mvt.
Therefore m (vtf vt) = ra^ (1 + e) vn
and vtf =Vtn(l+e) vn.
Substituting this value of vt' in the expression for tan which is obtained from Fig. 121, we get
vn evn
Eliminating vt between the last equation and the relation
tan a = ~ we obtain vn
e tan 0 = tan a - M (1 + e). (XV)
DISCUSSION. When AC = 0, equation (XV) reduces to equation (XIV).
When M = oo , tan /3= ooor/3 = -; therefore the particle slides along
2i
the plane towards the left. When e 0 and tan a > p, tan /3 = oo and ft = ~; therefore the particle slides along the plane towards the right.